Chapter 4, Problem 24SE

a.

To determine

Find the value of ${\mu }_X$.

Answer to Problem 24SE

The value of ${\mu }_X=\frac{p}{{\lambda }_1}+\frac{1-p}{{\lambda }_2}$.

Explanation of Solution

Given info:

Radioactive mass 1 and mass 2 emit particles at a mean rate of ${\lambda }_1\text{and }{\lambda }_2$ per second. The probability of selecting mass 1 is p and mass 2 is 1-p. The random variable X is defined as the time at which the first particle is emitted and it follows mixed exponential distribution with density function,

$f\left(x\right)=\{\begin{array}{l}p{\lambda }_1{e}^{{\lambda }_{1}x}+\left(1-p\right){\lambda }_2{e}^{-{\lambda }_{2}x}x\ge 0\\ 0x<0\end{array}$

Calculation:

For a continuous random variable with probability density function $f\left(x\right)$, the mean is,

${\mu }_X={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}xf\left(x\right)}dx$

Substitute $f\left(x\right)=p{\lambda }_1{e}^{{\lambda }_{1}x}+\left(1-p\right){\lambda }_2{e}^{-{\lambda }_{2}x}$ and integrate from 0 to ∞ in the above formula.

$\begin{array}{c}{\mu }_X={\displaystyle \underset{0}{\overset{\infty }{\int }}x\left(p{\lambda }_1{e}^{-{\lambda }_{1}x}+\left(1-p\right){\lambda }_2{e}^{-{\lambda }_{2}x}\right)}dx\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}\left(xp{\lambda }_1{e}^{-{\lambda }_{1}x}+x\left(1-p\right){\lambda }_2{e}^{-{\lambda }_{2}x}\right)}dx\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}\left(xp{\lambda }_1{e}^{-{\lambda }_{1}x}+x{\lambda }_2{e}^{-{\lambda }_{2}x}-px{\lambda }_2{e}^{-{\lambda }_{2}x}\right)}dx\\ =\left({\frac{xp{\lambda }_1{e}^{-{\lambda }_{1}x}}{-{\lambda }_1}|}_{x=0}^{x=\infty }-{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{p{\lambda }_1{e}^{-{\lambda }_{1}x}}{-{\lambda }_1}}\right)+\left({\frac{x{\lambda }_2{e}^{-{\lambda }_{2}x}}{-{\lambda }_2}|}_{x=0}^{x=\infty }-{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{{\lambda }_2{e}^{-{\lambda }_{2}x}}{-{\lambda }_2}}\right)-\left({\frac{xp{\lambda }_2{e}^{-{\lambda }_{2}x}}{-{\lambda }_2}|}_{x=0}^{x=\infty }-{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{p{\lambda }_2{e}^{-{\lambda }_{2}x}}{-{\lambda }_2}}\right)\end{array}$

      $\begin{array}{c}=\left(0+{\frac{p{e}^{-{\lambda }_{1}x}}{-{\lambda }_1}|}_{x=0}^{x=\infty }\right)+\left(0+{\frac{e^{-{\lambda }_{2}x}}{-{\lambda }_2}|}_{x=0}^{x=\infty }\right)-\left(0+{\frac{p{e}^{-{\lambda }_{2}x}}{-{\lambda }_2}|}_{x=0}^{x=\infty }\right)\\ =\frac{-p{e}^{-\infty }}{{\lambda }_1}+\frac{p{e}^0}{{\lambda }_1}+\frac{-e^{-\infty }}{{\lambda }_2}+\frac{e^0}{{\lambda }_2}+\frac{p{e}^{-\infty }}{{\lambda }_2}-\frac{p{e}^0}{{\lambda }_2}\\ =0+\frac{p}{{\lambda }_1}+0+\frac{1}{{\lambda }_2}+0-\frac{p}{{\lambda }_2}\\ =\frac{p}{{\lambda }_1}+\frac{1-p}{{\lambda }_2}\end{array}$

Thus, the value of mean is, ${\mu }_X=\frac{p}{{\lambda }_1}+\frac{1-p}{{\lambda }_2}$

b.

To determine

Find the cumulative distribution function of X.

Answer to Problem 24SE

The cumulative distribution function of X is, $F\left(x\right)=\{\begin{array}{l}1-p{e}^{-{\lambda }_{1}x}-\left(1-p\right)e^{-{\lambda }_{2}x}x\ge 0\\ 0x<0\end{array}$

Explanation of Solution

Calculation:

For a continuous random variable with probability density function $f\left(x\right)$, the cumulative distribution function can be obtained by integrating $f\left(x\right)$ in the range –∞ to x.

For $x<0$,

$\begin{array}{c}F\left(x\right)={\displaystyle \underset{-\infty }{\overset{x}{\int }}0}dt\\ =0\end{array}$

For $x\ge 0$

Substitute $f\left(x\right)=p{\lambda }_1{e}^{{\lambda }_{1}x}+\left(1-p\right){\lambda }_2{e}^{-{\lambda }_{2}x}$ and integrate from –∞ to x in the above formula.

$\begin{array}{c}F\left(x\right)={\displaystyle \underset{-\infty }{\overset{x}{\int }}p{\lambda }_1{e}^{{\lambda }_{1}t}+\left(1-p\right){\lambda }_2{e}^{-{\lambda }_{2}t}}dt\\ ={\displaystyle \underset{-\infty }{\overset{x}{\int }}\left(p{\lambda }_1{e}^{-{\lambda }_{1}t}+\left(1-p\right){\lambda }_2{e}^{-{\lambda }_{2}t}\right)}dt\\ ={\displaystyle \underset{-\infty }{\overset{x}{\int }}\left(p{\lambda }_1{e}^{-{\lambda }_{1}t}+{\lambda }_2{e}^{-{\lambda }_{2}t}-p{\lambda }_2{e}^{-{\lambda }_{2}t}\right)}dt\\ =\left({\frac{p{\lambda }_1{e}^{-{\lambda }_{1}t}}{-{\lambda }_1}|}_{t=-\infty }^{t=x}\right)+\left({\frac{{\lambda }_2{e}^{-{\lambda }_{2}t}}{-{\lambda }_2}|}_{t=-\infty }^{t=x}\right)-\left({\frac{p{\lambda }_2{e}^{-{\lambda }_{2}t}}{-{\lambda }_2}|}_{t=-\infty }^{t=x}\right)\end{array}$

      $\begin{array}{c}=-p{e}^{-{\lambda }_{1}x}+p{e}^0+-e^{-{\lambda }_{2}x}+e^0+p{e}^{-{\lambda }_{2}x}-p{e}^0\\ =-p{e}^{-x}+p+-e^{-{\lambda }_{2}x}+1+p{e}^{-{\lambda }_{2}x}-p\\ =1-p{e}^{-{\lambda }_{1}x}-\left(1-p\right)e^{-{\lambda }_{2}x}\end{array}$

Thus, the cumulative distribution function, $F\left(x\right)=\{\begin{array}{l}1-p{e}^{-{\lambda }_{1}x}-\left(1-p\right)e^{-{\lambda }_{2}x}x\ge 0\\ 0x<0\end{array}$

c.

To determine

Find the value of $P\left(X\le 2\right)$ for ${\lambda }_1=2,{\lambda }_2=1\text{and }p=0.5$.

Answer to Problem 24SE

The value of $P\left(X\le 2\right)$ is 0.9232.

Explanation of Solution

Calculation:

The required probability is $P\left(X\le 2\right)$. That is,

$P\left(X\le 2\right)=F\left(2\right)$

Substitute x as 2, ${\lambda }_1$ as 2, ${\lambda }_2$ as 1 and p as 0.5 in the formula of cumulative distribution function of X ,

$\begin{array}{c}P\left(X\le 2\right)=1-\left(0.5\right)e^{-2\left(2\right)}-\left(1-0.5\right)e^{-2\left(1\right)}\\ =1-0.5e^{-4}-0.5e^{-2}\\ =1-\left(0.5\right)\left(0.0183\right)-\left(0.5\right)\left(0.1353\right)\\ =1-0.00915-0.06765\\ =0.9232\end{array}$

Thus, the probability is $P\left(X\le 2\right)=0.9232$.

d.

To determine

Find the probability that mass 1 is selected if $P\left(X\le 2\right)$ is given.

Answer to Problem 24SE

The probability that mass 1 is selected is 0.5366.

Explanation of Solution

Given info:

Here ${\lambda }_1=2,{\lambda }_2=1\text{and }p=0.5$.

Calculation:

The random variable $M_1$ is defined as the event that mass 1 is selected.

The required probability is, $P\left(M_1|X\le 2\right)$.

Then,

$P\left(M_1|X\le 2\right)=\frac{P\left(X\le 2\cap M_1\right)}{P\left(X\le 2\right)}$

From part (c), it can be seen that, $P\left(X\le 2\right)=0.9232$

Now,

$\begin{array}{c}P\left(X\le 2\cap M_1\right)=P\left(X\le 2|M_1\right)P\left(M_1\right)\\ =\left(1-p{e}^{-{\lambda }_{1}x}\right)p\end{array}$

Substitute x as 2, ${\lambda }_1$ as 2 and p as 0.5 in the above formula

$\begin{array}{c}P\left(X\le 2\cap M_1\right)=\left(1-0.5e^{-2\left(2\right)}\right)\left(0.5\right)\\ =\left(1-\left(0.5\right)\left(0.0183\right)\right)\left(0.5\right)\\ =\left(1-0.00915\right)\left(0.5\right)\\ =\left(0.99085\right)\left(0.5\right)\\ =0.4954\end{array}$

Therefore,

$\begin{array}{c}P\left(M_1|X\le 2\right)=\frac{0.4954}{0.9232}\\ =0.5366\end{array}$

Thus, the probability that mass 1 is selected, if $P\left(X\le 2\right)$ is given is 0.5366.